If a= √3-√2/√3+√2 and b= √3+√2/√3-√2. Find the value od a²+b²?
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Solution:-
a = (√3-√2)/(√3+√2)
a = (√3-√2)/(√3+√2) × (√3-√2)/(√3-√2)
⇒ (3+2-2√3√2)/3-2
⇒ a = 5-2√6
b = (√3+√2)/(√3-√2)
b = (√3+√2)/(√3-√2) × (√3+√2)/(√3+√2)
⇒ (3+2+2√3√2)/3-2
b = 5+2√6
a² + b² = (5-2√6)² +(5+2√6)²
⇒ (25+24-20√6) + (25+24+20√6)
⇒ 25+24+25+24
= 98 Answer.
a = (√3-√2)/(√3+√2)
a = (√3-√2)/(√3+√2) × (√3-√2)/(√3-√2)
⇒ (3+2-2√3√2)/3-2
⇒ a = 5-2√6
b = (√3+√2)/(√3-√2)
b = (√3+√2)/(√3-√2) × (√3+√2)/(√3+√2)
⇒ (3+2+2√3√2)/3-2
b = 5+2√6
a² + b² = (5-2√6)² +(5+2√6)²
⇒ (25+24-20√6) + (25+24+20√6)
⇒ 25+24+25+24
= 98 Answer.
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